Poisson distribution

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My informal derivation feels too formal, but isn't formal enough to be a formal one! Work in progress!
Definition Poisson distribution [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] [ilmath]\lambda\in\mathbb{R}_{\ge 0} [/ilmath]([ilmath]\lambda[/ilmath] - the average rate of events per unit) file.png Discrete, over [ilmath]\mathbb{N}_{\ge 0} [/ilmath] $\mathbb{P}[X\eq k]:\eq e^{-\lambda}\frac{\lambda^k}{k!}$ $\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{i\eq 0}\frac{\lambda^i}{k!}$ [ilmath]\mathbb{E}[X]\eq\lambda[/ilmath] $2\lambda e^{-\lambda}\frac{\lambda^{u} }{u!}$[1]for: [ilmath]u:\eq[/ilmath][ilmath]\text{Floor} [/ilmath][ilmath](\lambda)[/ilmath] [ilmath]\text{Var}(X)\eq\lambda[/ilmath]

Definition

• [ilmath]X\sim\text{Poisson}(\lambda)[/ilmath]
• for [ilmath]k\in\mathbb{N}_{\ge 0} [/ilmath] we have: $\mathbb{P}[X\eq k]:\eq\frac{e^{-\lambda}\lambda^k}{k!}$
• the first 2 terms are easy to give: [ilmath]e^{\lambda} [/ilmath] and [ilmath]\lambda e^{-\lambda} [/ilmath] respectively, after that we have [ilmath]\frac{1}{2}\lambda^2 e^{-\lambda} [/ilmath] and so forth
• for [ilmath]k\in\mathbb{N}_{\ge 0} [/ilmath] we have: $\mathbb{P}[X\le k]\eq e^{-\lambda}\sum^k_{j\eq 0}\frac{1}{j!}\lambda^j$

As a formal random variable

 Situation for our RV [ilmath]\xymatrix{ & [0,1] \ar[r]^X & \mathbb{N}_0 \\ & \mathcal{B}([0,1]) \ar[dl]_-{\lambda} & \mathcal{P}(\mathbb{N}_0) \ar[l]_-{X^{-1} } \ar@{-->}@/^1em/[dll]^-{\mathbb{P}:\eq \lambda\circ X^{-1} } \\ \mathbb{R} & & } [/ilmath]
Caveat:[ilmath]\lambda[/ilmath] here is used to denote 2 things - the parameter to the Poisson distribution, and the restriction of the 1 dimensional Lebesgue measure to some region of interest.

There is no unique way to define a random variable, here is one way.

• Let [ilmath]\big([/ilmath][ilmath][0,1][/ilmath][ilmath],\ [/ilmath][ilmath]\mathcal{B}([0,1])[/ilmath][ilmath],\ [/ilmath][ilmath]\lambda[/ilmath][ilmath]\big)[/ilmath] be a probability space - which itself could be viewed as a rectangular distribution's random variable
• Let [ilmath]\lambda\in\mathbb{R}_{>0} [/ilmath] be given, and let [ilmath]X\sim\text{Poi}(\lambda)[/ilmath]
• Specifically consider [ilmath]\big(\mathbb{N}_0,\ [/ilmath][ilmath]\mathcal{P}(\mathbb{N}_0)[/ilmath][ilmath]\big)[/ilmath] as a sigma-algebra and [ilmath]X:[0,1]\rightarrow\mathbb{N}_0[/ilmath] by:
• [ilmath]X:x\mapsto\left\{\begin{array}{lr}0&\text{if }x\in[0,p_1)\\1 & \text{if }x\in[p_1,p_2)\\ \vdots & \vdots \\ k & \text{if }x\in[p_k,p_{k+1})\\ \vdots & \vdots \end{array}\right.[/ilmath] for $p_1:\eq e^{-\lambda} \frac{\lambda^1}{1!}$ and [ilmath]p_k:\eq p_{k-1}+e^{-\lambda}\frac{\lambda^k}{k!} [/ilmath]

Giving the setup shown on the left.

To add to page

Other

• Notes:Poisson and Gamma distribution
• I have proved (ages ago) and found a source that if we have [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] is the distribution of "observable events" and say that the probability of each event being observed is [ilmath]p[/ilmath], (that is observing an event is an i.i.d process, whether we observe or not being [ilmath]\sim[/ilmath][ilmath]\text{BORV} [/ilmath][ilmath](p)[/ilmath], with [ilmath]1[/ilmath] indicating 'observed' and [ilmath]0[/ilmath] indicating 'missed') then the distribution of observations, say [ilmath]Y[/ilmath] follows [ilmath]\sim\text{Poi}(p\lambda)[/ilmath]
• As with the above and "Poisson+Poisson=Poisson" we'd see that the Poisson distributions form some sort of algebraic structure.
• This is not to be confused with the definition of a Poisson mixture.

Mean

• $\sum^\infty_{n\eq 0} n\times\mathbb{P}[X\eq n]\eq\sum^\infty_{n\eq 0}\left[ n\times e^{-\lambda}\frac{\lambda^n}{n!}\right]\eq$$0+\left[ e^{-\lambda}\sum^\infty_{n\eq 1} \frac{\lambda^n}{(n-1)!}\right]$$\eq e^{-\lambda}\lambda\left[\sum^\infty_{n\eq 1}\frac{\lambda^{n-1} }{(n-1)!}\right]$
$\eq \lambda e^{-\lambda}\left[\sum^{\infty}_{n\eq 0}\frac{\lambda^n}{n!}\right]$$\eq \lambda e^{-\lambda}\left[\lim_{n\rightarrow\infty}\left(\sum^{n}_{k\eq 0}\frac{\lambda^k}{k!}\right)\right]$
• But! $e^x\eq\lim_{n\rightarrow\infty}\left(\sum^n_{i\eq 0}\frac{x^i}{i!}\right)$
• So $\eq\lambda e^{-\lambda} e^\lambda$
• [ilmath]\eq\lambda[/ilmath]

Mdm

See: Mdm of the Poisson distribution for proof.

For [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] we have:

• $\text{Mdm}(X)\eq 2\lambda e^{-\lambda}\frac{\lambda^u}{u!}$ for

Derivation of the Poisson distribution

Standard Poisson distribution:

• Let [ilmath]S:\eq[0,1)\subseteq\mathbb{R} [/ilmath], recall that means [ilmath]S\eq\{x\in\mathbb{R}\ \vert\ 0\le x<1\} [/ilmath]
• Let [ilmath]\lambda[/ilmath] be the average count of some event that can occur [ilmath]0[/ilmath] or more times on [ilmath]S[/ilmath]

We will now divide [ilmath]S[/ilmath] up into [ilmath]N[/ilmath] equally sized chunks, for [ilmath]N\in\mathbb{N}_{\ge 1} [/ilmath]

• Let [ilmath]S_{i,N}:\eq\left[\frac{i-1}{N},\frac{i}{N}\right)[/ilmath][Note 1] for [ilmath]i\in\{1,\ldots,N\}\subseteq\mathbb{N} [/ilmath]

We will now define a random variable that counts the occurrences of events per interval.

• Let [ilmath]C\big(S_{i,N}\big)[/ilmath] be the RV such that its value is the number of times the event occurred in the [ilmath]\left[\frac{i-1}{N},\frac{i}{N}\right)[/ilmath] interval

We now require:

• $\lim_{N\rightarrow\infty}\left(\mathbb{P}[C\big(S_{i,N}\big)\ge 2]\right)\eq 0$ - such that:
• as the [ilmath]S_{i,N} [/ilmath] get smaller the chance of 2 or more events occurring in the space reaches zero.
• Warning:This is phrased as a limit, I'm not sure it should be as we don't have any [ilmath]S_{i,\infty} [/ilmath] so no [ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath] distribution then either

Note that:

• $\lim_{N\rightarrow\infty}\big(C(S_{i,N})\big)\eq\lim_{N\rightarrow\infty}\left(\text{BORV}\left(\frac{\lambda}{N}\right)\right)$
• This is supposed to convey that the distribution of [ilmath]C(S_{i,N})[/ilmath] as [ilmath]N[/ilmath] gets large gets arbitrarily close to [ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath]

So we may say for sufficiently large [ilmath]N[/ilmath] that:

• $C(S_{i,N})\mathop{\sim}_{\text{(approx)} }$[ilmath]\text{BORV}(\frac{\lambda}{N})[/ilmath], so that:
• [ilmath]\mathbb{P}[C(S_{i,N})\eq 0]\approx(1-\frac{\lambda}{N}) [/ilmath]
• [ilmath]\mathbb{P}[C(S_{i,N})\eq 1]\approx \frac{\lambda}{N} [/ilmath], and of course
• [ilmath]\mathbb{P}[C(S_{i,N})\ge 2]\approx 0[/ilmath]

Assuming the [ilmath]C(S_{i,N})[/ilmath] are independent over [ilmath]i[/ilmath] (which surely we get from the [ilmath]\text{BORV} [/ilmath] distributions?) we see:

• $C(S)\mathop{\sim}_{\text{(approx)} }$[ilmath]\text{Bin} [/ilmath]$\left(N,\frac{\lambda}{N}\right)$ or, more specifically: $C(S)\eq\lim_{N\rightarrow\infty}\Big(\sum^N_{i\eq 1}C(S_{i,N})\Big)\eq\lim_{N\rightarrow\infty}\left(\text{Bin}\left(N,\frac{\lambda}{N}\right)\right)$

We see:

• $\mathbb{P}[C(S)\eq k]\eq\lim_{N\rightarrow\infty}$[ilmath]\Big(\mathbb{P}\big[\text{Bin}(N,\frac{\lambda}{N})\eq k\big]\Big)[/ilmath]$\eq\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)$

We claim that:

• $\lim_{N\rightarrow\infty}\left({}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k\left(1-\frac{\lambda}{N}\right)^{N-k}\right)\eq \frac{\lambda^k}{k!}e^{-\lambda}$

We will tackle this in two parts:

• $\lim_{N\rightarrow\infty}\Bigg(\underbrace{ {}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k}_{A}\ \underbrace{\left(1-\frac{\lambda}{N}\right)^{N-k} }_{B}\Bigg)$ where [ilmath]B\rightarrow e^{-\lambda} [/ilmath] and $A\rightarrow \frac{\lambda^k}{k!}$

Proof

Key notes:

A

Notice:

• ${}^N\!C_k\ \left(\frac{\lambda}{N}\right)^k \eq \frac{N!}{(N-k)!k!}\cdot\frac{1}{N^k}\cdot\lambda^k$
$\eq\frac{1}{k!}\cdot\frac{\overbrace{N(N-1)\cdots(N-k+2)(N-k+1)}^{k\text{ terms} } }{\underbrace{N\cdot N\cdots N}_{k\text{ times} } } \cdot\lambda^k$
• Notice that as [ilmath]N[/ilmath] gets bigger [ilmath]N-k+1[/ilmath] is "basically" [ilmath]N[/ilmath] so the [ilmath]N[/ilmath]s in the denominator cancel (in fact the value will be slightly less than 1, tending towards 1 as [ilmath]N\rightarrow\infty[/ilmath]) this giving:
• $\frac{\lambda^k}{k!}$

B

This comes from:

• $e^x:\eq\lim_{n\rightarrow\infty}\left(\left(1+\frac{x}{n}\right)^n\right)$, so we get the [ilmath]e^{-\lambda} [/ilmath] term.

Notes

1. Recall again that means [ilmath]\{x\in\mathbb{R}\ \vert\ \frac{i-1}{N}\le x < \frac{i}{N} \} [/ilmath]